Consider the function $f(x) = \begin{cases} x \sin \frac{\pi}{x} & \text{for } x > 0 \\ 0 & \text{for } x = 0 \end{cases}$. Then the number of points in $(0, 1)$ where the derivative $f'(x)$ vanishes is:

Let $f(x) = 8x^3 - 6x^2 - 2x + 1,$ then

If the function $f(x) = x^3 - 6ax^2 + 5x$ satisfies the conditions of Lagrange's mean value theorem for the interval $[1, 2]$ and the tangent to the curve $y = f(x)$ at $x = \frac{7}{4}$ is parallel to the chord that joins the points of intersection of the curve with the ordinates $x = 1$ and $x = 2$,then the value of $a$ is

If $f(x) = \begin{cases} x, & 0 \leq x \leq 1 \\ 2-x, & 1 < x \leq 2 \end{cases}$, then Rolle's theorem is not applicable to $f(x)$ because

The value of $c$ in the Lagrange's mean value theorem for $f(x)=\sqrt{x-2}$ in the interval $[2,6]$ is

Verify the Mean Value Theorem for the function $f(x) = x^{2} - 4x - 3$ in the interval $[a, b]$,where $a = 1$ and $b = 4$.